Using the position function -16t^2=1000

and

Lim  s(a)-s(t)
t->a    a-t

If a construction wokrer drops a wrench from a height of 1000ft, when
will the wrench hit the ground? At what velocity will the wrench
impact the ground?

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Clarification of Question by yalexa10-ga on 09 Jul 2006 12:30 PDT

the position function is 16t^2-1000 and the limit is s(a)-s(t) / a-t

Hi yalexa10-ga,

At the beginning, when t = 0, the position of the wrench is 1000ft
above ground. Therefore, the position function would be:

   s=1000-16t^2   (where 's' is feet above ground level)

At the moment of impact, s=0, which gives:

   0=1000-16t^2

Rearranging this gives:

   16t^2=1000

then:

   t^2=1000/16

then:

   t=(1000/16)^0.5

In other words, t equals the square root of 1000/16, or 7.90569 seconds.

To find the velocity at impact we consider the velocity to be the
limit (as 't' approaches the time of impact 'a') of:

   v = ( s(a) - s(t) ) / (a-t)

Substituting the position function gives:

   v = ( 1000-16a^2 - (1000-16t^2) ) / (a-t)

which is equivalent to:

   v = (1000 - 16a^2 - 1000 + 16t^2) / (a-t)

which is equivalent to:

   v = 16(t^2-a^2)/(a-t)

It is an algebraic identity, true for any 't' and 'a', that t^2-a^2 is
equal to (t+a)(t-a), which is equal to -(a+t)(a-t). Substituting this
gives:

   v = 16(a+t)(a-t)/(a-t)

The two occurrences of (a-t) cancel out, leaving us with:

   v = 16(a+t)

In the limit, as 't' approaches the time of impact 'a', this is equivalent to

   v = 16(a+a)

Our time of impact is 7.90569 seconds, therefore the velocity at impact must be:

   16(7.90569+7.90569)

which is 252.982 feet per second.

I trust this provides the information you require. If any of the steps
are not clear, please request clarification.

Regards,
eiffel-ga


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